The vertices of a triangle are `[a t_1t_2,a(t_1 +t_2)]`, `[a t_2t_3,a(t_2 +t_3)]`, `[a t_3t_1,a(t_3 +t_1)]` Then the orthocenter of the triangle is (a) `(-a, a(t_1+t_2+t_3)-at_1t_2t_3)` (b) `(-a, a(t_1+t_2+t_3)+at_1t_2t_3)` (c) `(a, a(t_1+t_2+t_3)+at_1t_2t_3)` (d) `(a, a(t_1+t_2+t_3)-at_1t_2t_3)`

Answer

`(-a,a(t_1+t_2+t_3)+at_1t_2t_3)`

Explanation

We know that orthocenter of `Delta` is point of concurrencyof altitudes. Let orthocenter be `H(x,y)`.
Now, slope of `BC=((t_1+t_3)-a(t_2+t_3))/(at_1t_3-at_2t_3)`
`BC=((t_1+t_3-t_2-t_3))/(at_1(t_1-t_2))=(1)/(t_3)`
`therefore` Slope of` AD =-t_3`
Also, slope of AD=Slope of `AH=(y-(t_1+t_2))/(x-at_1t_2)=-t_3`
or `xt_3=at_1t_2t_3+a(t_1+t_2)`
Similarly , by symmetry form `AC botBE`, we get (1)
`xt_1+y=at_1t_2t_3+a(t_2_t_3)` (2)
Solving (1) and (2),we get
`x=-a,y=a(t_1+t_2+t_3_+at_1t_2t_3`
`therefore` Orthocenter is `H(-a,a(t_1+t_2+t_3)+at_1t_2t_3)`